bonanova

I think it's solved. Each of these were mentioned on one post or another: strengths strengths, stretched queueing [angsts] witchcraft lengths strengths latchstring [hyphen is optional] facetiously abstemiously indivisibility [indivisibilities if you want to show off - it has7 occurrences of i] bookkeeper

Very nice.!

OK so there is this black velvet bag that contains a bunch of uniquely numbered balls spoons. Every puzzle has numbered balls. Enough with the balls. This puzzle has spoons. You will be asked two questions about these spoons. In order to get the needed information I ask my beautiful assistant here [pic load failed; sorry ] to remove from the bag some quantity of spoons, add their numbers and write it on a sheet of paper. She looks at you, smiles, and suggests that should be enough information for you. You disagree and ask her to replace the spoons and repeat the process, which she does, drawing the same number of spoons from the bag but this time the sum she writes is a different number. You ask her to do it again. And again, ... until the sums begin to repeat themselves, and eventually you are convinced that no more new numbers will appear. She then erases any duplicates, keeping only one occurrence of each, and hands you the paper. On it you see: 115 118 113 110 120 117 116 112 121 114 Then you are asked: How many spoons are in the bag, and what are their numbers? You think for only a moment and say that you need more information, like how many spoons were drawn each time. My smiling assistant says she can't tell you that; but she can tell you that the sum of the number of spoons in the bag and the number of spoons drawn each time is an odd number. You pull out a used envelope, turn it over [everyone does this], scribble something, and then give your answer.

I have 12 sticks and I have painted 8 of them red and 4 of them blue. I then glued the sticks together to form the frame of a cube. So now I'm wondering how many total ways could I have assembled the sticks into cubes that could be distinguished from each other. Assume I can pick up the cube and move it around any way I like, but I cannot deform or re-assemble it

I meant i didn't understand how you found out the answer using the number of nodes.How did you find the rectangles that are combinations of two or more rectangles using the number of nodes?Is there some formula or you just simply counted them? Just counted them. Using nodes just put them into piles sort of. A convenient way to keep track.

Bushindo, Thanks. I learn so much from you. Is there also a maximum likelihood estimate of the underlying assumptions that would make the maximum likelihood estimate of n well defined?

In the second question does the 1-inch spacing refer to center-center distance? Or must the edge of a stamp's disc be > 1 in from the center of any previous stamp? I think you mean the latter, but just to be certain, I'm asking. Meantime, here's the ...

You are both correct. nakulendu posted first, so I marked his answer as the solution.

It seems like this problem is dependent upon what is a reasonable a priori distribution for N. Oh wait, the answer is much simpler than that. Turns out this is the easiest bonanova puzzle I have ever seen =) LOL That may be the first and only time I use LOL in this forum. I'm not that clever. Really. But I love it. It's a better puzzle than the one I intended. Honorable mention. .

You're right about the bisector case. But for the trisector case there is more than one point. In fact there are three places where a trisector of one angle first intersects a trisector of one of the other angles. And there is something special about those three points.

By nodes I just meant corners: places where the lines intersect. (In graph theory, those points are called nodes, and lines that connect them are called edges.)

Since you don't know the value of n, but are asked to estimate it, you don't know that p and n are different.

Suppose n tickets numbered 1-n are in a box, and you draw one of them You don't know how many tickets are in the box, but you are asked to estimate how many there are. Your ticket has the number p on it. What estimate of n has the highest likelihood of being correct?

Hi Krishna, I didn't mean to ignore your previous post. No, that's not the symbol I had in mind.

Warm-up problem: Bisect the angles of a triangle. Describe the point(s) where each bisector first intersects one of the others. Now try this: Trisect the angles of a triangle. Describe the points where each trisector first intersects one of the others.

TC, Thanks. I did not fully appreciate the "timing" aspect.

If I understand, a ticket has the numerical string 'abcdef', where a-f are in the set { 0 1 2 3 4 5 6 7 8 9 }, a+b+c = d+e+f duplicates are allowed. To re-state Prime's question, I will ask: is 000000 a lucky ticket? When you say "all the lucky ticket numbers" sum to a multiple of 13, which do you mean? for each and every lucky ticket abcdef, a+b+c+d+e+f is divisible by 13, or taking all possible lucky numbers, their sum, abcdef + ghijkl + ... + uvwxyz is divisible by 13

Shamelessly bumping this puzzle. It's worth the effort to solve, IMO. 12 trees making 18 rows of 3 trees each.

dm92 your answer answer agrees with that of several others. Y-san and I get a slightly smaller number. It puzzles me (npi) what is the difference in solving this that leads to two different solutions. Maybe the wording of the OP is ambiguous.

dms, I'm having difficulty seeing the extra ones. Although, the ones I found seemed straightforward, making me think there are others not readily seen. Any rectangle has to have an upper left corner, which therefore can't be any of the bottom or right nodes. Using the order that I gave, what is your count for the six other nodes? That will help me see the ones I missed.